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Theorem ddif 4097
Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)
Assertion
Ref Expression
ddif (V ∖ (V ∖ 𝐴)) = 𝐴

Proof of Theorem ddif
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 vex 3448 . . . . 5 𝑥 ∈ V
2 eldif 3921 . . . . 5 (𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥𝐴))
31, 2mpbiran 708 . . . 4 (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥𝐴)
43con2bii 358 . . 3 (𝑥𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴))
51biantrur 532 . . 3 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)))
64, 5bitr2i 276 . 2 ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥𝐴)
76difeqri 4085 1 (V ∖ (V ∖ 𝐴)) = 𝐴
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wa 397   = wceq 1542  wcel 2107  Vcvv 3444  cdif 3908
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-v 3446  df-dif 3914
This theorem is referenced by:  complss  4107  dfun3  4226  dfin3  4227  invdif  4229  ssindif0  4424  difdifdir  4450
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